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2 years ago

In a nuclear reactor a moderator is used to

1 answer:

4 0

Answer: A material called moderator is used to slow down the neutrons released during nuclear fission. The source's neutrons have a high speed and energy. The neutrons' speed is slowed by heavy water or graphite moderators.

Explanation:

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Plants and trees grow nearly everywhere. this is one of the advantages of what energy?

AnnyKZ [126]

The answer is Biomass.

Letter :D

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3 years ago

Read 2 more answers

A dragster and driver together have mass 931.9 kg . The dragster, starting from rest, attains a speed of 25.7 m/s in 0.57 s. Fin

Mama L [17]

Answer:

45.09 m/s^2

Explanation:

from the question we were given the following parameters

mass = 931.9 kg

initial velocity (V1) = 0 m/s ( since it started from rest )

final velocity (V2) = 25.7 m/s

time (t) = 0.57 s

The average acceleration is given by:

a = ( V2 - V1 ) / t

therefore

a = (25.7 - 0) / 0.57 = 45.09 m/s^2

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3 years ago

What type of friction prevents a pile of rocks from falling apart?

BlackZzzverrR [31]

A because it some type of friction

6 0

3 years ago

Which equation can be used to model simple harmonic motion

mixer [17]

Answer:x(t)= Acos(wt)

Explanation:

According to Newton's 2nd law,a particle of mass m acted on by a force is given by:Fs=-kx

Where x is displacement from equilibrium

K = spring constant

Therefore X(t) = Acos(2pit/T)

X(t)= Acos(wt)

8 0

3 years ago

An adiabatic nozzle has an inlet area of 1 m^2 and an outlet area of 0.25 m^2. Water enters the nozzle at a rate of 5 m^3/s and

svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

$V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s$

velocity at the exit=20m/s

for entry

$V=\frac{5}{1} =5m/s$

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

$\frac{P1}{\alpha } +\frac{v1^2}{2g} =\frac{P2} {\alpha } +\frac{v2^2}{2g}$

where

P=presure

α=9.810KN/m^3 specific weight for water

V=speed

g=gravity

solving for P1

$(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81 =p2\\P2=-37.5kPa$

the pressure at exit is -37.5kPa

7 0

4 years ago